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Why is the psuedoscalar in 5 dimensions the identity matrix?

In 3 dimensions its $\begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix}$

In 3 dimensions when you multiply all 3 vectors (Pauli matrices) together you get the unit psuedoscalar. Multiplying a vector (Pauli matrix) times the psuedoscalar results in a bivector (quaternion)

$ \begin{array}{r|rrr|rrr|r} {\color{red} {I_2}} & {\color{green} {\sigma_1}} & {\color{green} {\sigma_2}} & {\color{green} {\sigma_3}} & {\color{blue} {\boldsymbol{\hat{\imath}}}} & {\color{blue} {\boldsymbol{\hat{\jmath}}}} & {\color{blue} {\boldsymbol{\hat{k}}}} & {\color{red} {i_2}} \\\hline {\color{green} {\sigma_1}} & {\color{red} {I_2}} & {\color{blue} {\boldsymbol{\hat{\imath}}}} & {\color{blue} {\boldsymbol{-\hat{\jmath}}}} & {\color{green} {\sigma_2}} & {\color{green} {-\sigma_3}} & {\color{red} {i_2}} & {\color{blue} {\boldsymbol{\hat{k}}}} \\ {\color{green} {\sigma_2}} & {\color{blue} {\boldsymbol{-\hat{\imath}}}} & {\color{red} {I_2}} & {\color{blue} {\boldsymbol{\hat{k}}}} & {\color{green} {-\sigma_1}} & {\color{red} {i_2}} & {\color{green} {\sigma_3}} & {\color{blue} {\boldsymbol{\hat{\jmath}}}} \\ {\color{green} {\sigma_3}} & {\color{blue} {\boldsymbol{\hat{\jmath}}}} & {\color{blue} {\boldsymbol{-\hat{k}}}} & {\color{red} {I_2}} & {\color{red} {i_2}} & {\color{green} {\sigma_1}} & {\color{green} {-\sigma_2}} & {\color{blue} {\boldsymbol{\hat{\imath}}}} \\\hline {\color{blue} {\boldsymbol{\hat{\imath}}}} & {\color{green} {-\sigma_2}} & {\color{green} {\sigma_1}} & {\color{red} {i_2}} & {\color{red} {-I_2}} & {\color{blue} {\boldsymbol{\hat{k}}}} & {\color{blue} {\boldsymbol{-\hat{\jmath}}}} & {\color{green} {-\sigma_3}} \\ {\color{blue} {\boldsymbol{\hat{\jmath}}}} & {\color{green} {\sigma_3}} & {\color{red} {i_2}} & {\color{green} {-\sigma_1}} & {\color{blue} {\boldsymbol{-\hat{k}}}} & {\color{red} {-I_2}} & {\color{blue} {\boldsymbol{\hat{\imath}}}} & {\color{green} {-\sigma_2}} \\ {\color{blue} {\boldsymbol{\hat{k}}}} & {\color{red} {i_2}} & {\color{green} {-\sigma_3}} & {\color{green} {\sigma_2}} & {\color{blue} {\boldsymbol{\hat{\jmath}}}} & {\color{blue} {\boldsymbol{-\hat{\imath}}}} & {\color{red} {-I_2}} & {\color{green} {-\sigma_1}} \\\hline {\color{red} {i_2}} & {\color{blue} {\boldsymbol{\hat{k}}}} & {\color{blue} {\boldsymbol{\hat{\jmath}}}} & {\color{blue} {\boldsymbol{\hat{\imath}}}} & {\color{green} {-\sigma_3}} & {\color{green} {-\sigma_2}} & {\color{green} {-\sigma_1}} & {\color{red} {I_2}} \end{array} $

Identity matrix (scalar) is in red.

The Pauli matrices (vectors) are in green.

Quaternions (bivectors) are in blue.

Unit pseudoscalar is in red.

But when you go to 5 dimensions and use the Dirac matrices you get something totally different.

Multiplying any 2 vectors (Dirac matrices) results in a bivector as expected. But multiplying a bivector times a Dirac matrix results not in a trivector as expected but rather in another bivector.

And multiplying any 4 Dirac matrices results not in a quadvector as expeted but rather in a Dirac matrix (a vector).

Multiplying all 5 Dirac matrices result in the identity matrix instead of the pseudoscalar.

You can check my work here: https://math.wikia.com/wiki/Dirac_matrices#Alpha_multiplication_table and https://math.wikia.com/wiki/Pauli_matrices

Might take a minute to load though

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1 Answer 1

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https://en.wikipedia.org/wiki/Classification_of_Clifford_algebras#Unit_pseudoscalar

Apparently it's periodic. In seven Dimensions the pseudoscalar would be $i \cdot I$ again

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